1. Chapter #3
Matter and Energy

2. Matter Anything occupying space and having mass.
Matter exists in three states.
Solid
Liquid
Gas
Plasma

3. Different Physical Property
Mixture Separation
Mixtures can be separated based on different physical properties of the components.
Evaporation
Volatility
Chromatography
Adherence to a surface
Filtration
State of matter
(solid/liquid/gas)
Distillation
Boiling point
Technique
Different Physical Property

4. Mixture Separation by Boiling Point
Distillation of a Solution Consisting of Salt Dissolved in Water

5. Mixture Separation by State of Matter
Called Filtration
Separates a liquid from a solid.

6. Mixture Separation by Adherence to a surface
This method is called Chromatography
sand

7. Law of Conservation of Mass
Antoine Lavoisier
“Matter is neither created nor destroyed in a chemical reaction.”
The total amount of matter present before a chemical reaction is always the same as the total amount after.
The total mass of all the reactants is equal to the total mass of all the products.

8. Conservation of Mass 266 grams = 266 grams
Total amount of matter remains constant in a chemical reaction.
58 grams of butane burns in 208 grams of oxygen to form 176 grams of carbon dioxide and 90 grams of water.
butane oxygen  carbon dioxide + water
58 grams grams  grams grams
266 grams = grams

9. Energy There are things that do not have mass and volume.
These things fall into a category we call energy.
Energy is anything that has the capacity to do work.
Although chemistry is the study of matter, matter is effected by energy.
It can cause physical and/or chemical changes in matter.

10. Law of Conservation of Energy
“Energy can neither be created nor destroyed.”
The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount.
However, we can transfer energy from one place in the universe to another, and we can change its form.

11. Matter Possesses Energy
When a piece of matter possesses energy, it can give some or all of it to another object.
It can do work on the other object.
All chemical and physical changes result in the matter changing energy.

12. Kinetic and Potential Energy
Potential energy is energy that is stored.
Water flows because gravity pulls it downstream.
However, the dam won’t allow it to move, so it has to store that energy.
Kinetic energy is energy of motion, or energy that is being transferred from one object to another.
When the water flows over the dam, some of its potential energy is converted to kinetic energy of motion.

13. Forms of Energy Electrical Heat or Thermal Energy
Kinetic energy associated with the flow of electrical charge.
Heat or Thermal Energy
Kinetic energy associated with molecular motion.
Light or Radiant Energy
Kinetic energy associated with energy transitions in an atom.
Nuclear
Potential energy in the nucleus of atoms.
Chemical
Potential energy in the attachment of atoms or because of their position.

14. Converting Forms of Energy
When water flows over the dam, some of its potential energy is converted to kinetic energy.
Some of the energy is stored in the water because it is at a higher elevation than the surroundings.
The movement of the water is kinetic energy.
Along the way, some of that energy can be used to push a turbine to generate electricity.
Electricity is one form of kinetic energy.
The electricity can then be used in your home. For example, you can use it to heat cake batter you mixed, causing it to change chemically and storing some of the energy in the new molecules that are made.

15. Using Energy
We use energy to accomplish all kinds of processes, but according to the Law of Conservation of Energy we don’t really use it up!
When we use energy we are changing it from one form to another.
For example, converting the chemical energy in gasoline into mechanical energy to make your car move.

16. “Losing” Energy
If a process was 100% efficient, we could theoretically get all the energy transformed into a useful form.
Unfortunately we cannot get a 100% efficient process.
The energy “lost” in the process is energy transformed into a form we cannot use.

17. There’s No Such Thing as a Free Ride
When you drive your car, some of the chemical potential energy stored in the gasoline is released.
Most of the energy released in the combustion of gasoline is transformed into sound or heat energy that adds energy to the air rather than move your car down the road.

18. Units of Energy
Calorie (cal) is the amount of energy needed to raise one gram of water by 1 °C.
kcal = energy needed to raise 1000 g of water 1 °C.
food calories = kcals.
Energy Conversion Factors
1 calorie (cal) =
4.184 joules (J)
1 Calorie (Cal)
1000 calories (cal)
1 kilowatt-hour (kWh)
3.60 x 106 joules (J)

19. Energy Use Unit joule (J) 4.18 3.6 x 105 9.0 x 108 calorie (cal) 1.00
Energy Required to Raise Temperature of 1 g of Water by 1°C
Energy Required to Light 100-W Bulb for 1 Hour
Energy Used by Average U.S. Citizen in 1 Day
joule (J)
4.18
3.6 x 105
9.0 x 108
calorie (cal)
1.00
8.60 x 104
2.2 x 108
Calorie (Cal)
1.00 x 10-3
86.0
2.2 x 105
kWh
1.1 x 10-6
0.100
2.50 x 102

20. Chemical Potential Energy
The amount of energy stored in a material is its chemical potential energy.
The stored energy arises mainly from the attachments between atoms in the molecules and the attractive forces between molecules.
When materials undergo a physical change, the attractions between molecules change as their position changes, resulting in a change in the amount of chemical potential energy.
When materials undergo a chemical change, the structures of the molecules change, resulting in a change in the amount of chemical potential energy.

21. Energy Changes in Reactions
Chemical reactions happen most readily when energy is released during the reaction.
Molecules with lots of chemical potential energy are less stable than ones with less chemical potential energy.
Energy will be released when the reactants have more chemical potential energy than the products.

22. Exothermic Processes
When a change results in the release of energy it is called an exothermic process.
An exothermic chemical reaction occurs when the reactants have more chemical potential energy than the products.
The excess energy is released into the surrounding materials, adding energy to them.
Often the surrounding materials get hotter from the energy released by the reaction.

23. An Exothermic Reaction
Surroundings
reaction
Potential energy
Reactants
Products
Amount of energy released

24. Endothermic Processes
When a change requires the absorption of energy it is called an endothermic process.
An endothermic chemical reaction occurs when the products have more chemical potential energy than the reactants.
The required energy is absorbed from the surrounding materials, taking energy from them.
Often the surrounding materials get colder due to the energy being removed by the reaction.

25. An Endothermic Reaction
Surroundings
reaction
Potential energy
Products
Reactants
Amount of energy absorbed

26. Energy and the Temperature of Matter
The amount the temperature of an object increases depends on the amount of heat energy added (q).
If you double the added heat energy the temperature will increase twice as much.
The amount the temperature of an object increases depending on its mass.
If you double the mass, it will take twice as much heat energy to raise the temperature the same amount.

27. Heat Capacity
Heat capacity is the amount of heat a substance must absorb to raise its temperature by 1 °C.
cal/°C or J/°C.
Metals have low heat capacities; insulators have high heat capacities.
Specific heat = heat capacity of 1 gram of the substance.
cal/g°C or J/g°C.
Water’s specific heat = J/g°C for liquid.
Or cal/g°C.
It is less for ice and steam.

28. Specific Heat Capacity
Specific heat is the amount of energy required to raise the temperature of one gram of a substance by 1 °C.
The larger a material’s specific heat is, the more energy it takes to raise its temperature a given amount.
Like density, specific heat is a property of the type of matter.
It doesn’t matter how much material you have.
It can be used to identify the type of matter.
Water’s high specific heat is the reason it is such a good cooling agent.
It absorbs a lot of heat for a relatively small mass.

29. Specific Heat Capacities

30. Heat Gain or Loss by an Object
The amount of heat energy gained or lost by an object depends on 3 factors: how much material there is, what the material is, and how much the temperature changed.

31. Practice—Calculate the Amount of Heat Released When 7
Practice—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C

32. Practice—Calculate the Amount of Heat Released When 7
Practice—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water j/g-C and cross out all units except the heat unit, the joule (j), using our four step process
4.184 j
g- °C

33. Practice—Calculate the Amount of Heat Released When 7
Practice—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water j/g-C and cross out all units except the heat unit, the joule (j), using our four step process
4.184 j
7.40 g
g- °C

34. Practice—Calculate the Amount of Heat Released When 7
Practice—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water j/g-C and cross out all units except the heat unit, the joule (j), using our four step process
4.184 j
7.40 g
20 °C
g- °C

35. Practice—Calculate the Amount of Heat Released When 7
Practice—Calculate the Amount of Heat Released When 7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water j/g-C and cross out all units except the heat unit, the joule (j), using our four step process
4.184 j
7.40 g
20 °C
= 620 j
g- °C

36. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity?

37. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C

38. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C
Now we will organize the information in the problem to give the required units.

39. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C
Now we will organize the information in the problem to give the required units.
442 j

40. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C
Now we will organize the information in the problem to give the required units.
442 j
3.22 g

41. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C
Now we will organize the information in the problem to give the required units.
442 j
3.22 g
(57.0 – 20.0) °C

42. Heat Problems
Calculate the specific heat capacity of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat capacity? j/g-°C
Now we will organize the information in the problem to give the required units.
442 j
=3.71j/g- °C
3.22 g
(57.0 – 20.0) °C

43. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.

44. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.

45. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.0 cal
g-°C

46. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.0 cal
Cal
g-°C
103cal

47. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.0 cal
Cal
88.6 g
g-°C
103cal

48. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.0 cal
Cal
88.6 g
( ) °C
g-°C
103cal

49. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.0 cal
Cal
88.6 g
( ) °C
g-°C
103cal
1.22 g

50. Heat Problems
Calculate the specific heat of a 3.22 g sample of metal, if it absorbs 442 j of heat and its temperature rises from 20.0°C to 57.0°C.
Find the food Calories, per gram, of a potato chip if a 1.22 g sample of chip causes 88.6 g of water to rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will organize the units to give Calories.
1.00 cal
Cal
88.6 g
( ) °C
= 5.59 Cal/g
g-°C
103cal
1.22 g

51. Calories and Exercise Energy Expended (Cal/h) Activity by 150 lb Adult
Sleeping 80
Sitting 100
Walking (2.5 mph) 324
Cycling (5.5 mph) 330
Skiing (downhill) 486
Basketball 564
Swimming (fast crawl) 636
1.0 lb body fat = 3500 Cal
1.0 Pt of ice cream = 6.00 X 102 Cal

52. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?

53. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr

54. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr Hr
324 Cal

55. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal Hr
324 Cal pt

56. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal
1.0 pt Hr
324 Cal pt

57. Sample Problem
How far must one walk at 2.5 mi/hr to burn off the Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal
1.0 pt
= 4.6 mi. Hr
324 Cal pt

58. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?

59. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?
2.5 mi hr

60. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?
2.5 mi hr hr
324 Cal

61. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?
2.5 mi hr
3500 Cal hr
324 Cal Lb

62. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?
2.5 mi hr
3500 Cal
1.0 Lb hr
324 Cal Lb

63. Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb of body fat?
2.5 mi hr
3500 Cal
1.0 Lb
= 27 mi hr
324 Cal Lb

64. The End